Linear Algebra · Worked Example 5
Find a Basis for the Space Spanned by Four Vectors
Problem 5. Find a basis for the space spanned by the vectors
\( v_1 = (1,-2,0,0,3) \), \( v_2 = (2,-5,-3,-2,6) \), \( v_3 = (0,5,15,10,0) \),
and \( v_4 = (2,6,18,8,6) \).
Solution
Write the vectors as the rows of a matrix:
\[ A = \begin{bmatrix} 1 & -2 & 0 & 0 & 3 \\ 2 & -5 & -3 & -2 & 6 \\ 0 & 5 & 15 & 10 & 0 \\ 2 & 6 & 18 & 8 & 6 \end{bmatrix} \]Step 1 — Row reduce A
\[
\begin{bmatrix}
1 & -2 & 0 & 0 & 3 \\
2 & -5 & -3 & -2 & 6 \\
0 & 5 & 15 & 10 & 0 \\
2 & 6 & 18 & 8 & 6
\end{bmatrix}
\]
→
\( R_2 \to R_2 - 2R_1 \)
\( R_4 \to R_4 - 2R_1 \)
\[
\begin{bmatrix}
1 & -2 & 0 & 0 & 3 \\
0 & -1 & -3 & -2 & 0 \\
0 & 5 & 15 & 10 & 0 \\
0 & 10 & 18 & 8 & 0
\end{bmatrix}
\]
→
\( R_2 \to -R_2 \)
\[
\begin{bmatrix}
1 & -2 & 0 & 0 & 3 \\
0 & 1 & 3 & 2 & 0 \\
0 & 5 & 15 & 10 & 0 \\
0 & 10 & 18 & 8 & 0
\end{bmatrix}
\]
→
\( R_1 \to R_1 + 2R_2 \)
\( R_3 \to R_3 - 5R_2 \)
\( R_4 \to R_4 - 10R_2 \)
\[
\begin{bmatrix}
1 & 0 & 0 & -2 & 3 \\
0 & 1 & 3 & 2 & 0 \\
0 & 0 & 0 & 0 & 0 \\
0 & 0 & -12 & -12 & 0
\end{bmatrix}
\]
→
\( R_3 \leftrightarrow R_4 \)
\[
\begin{bmatrix}
1 & 0 & 0 & -2 & 3 \\
0 & 1 & 3 & 2 & 0 \\
0 & 0 & -12 & -12 & 0 \\
0 & 0 & 0 & 0 & 0
\end{bmatrix}
\]
→
\( R_3 \to \dfrac{R_3}{-12} \)
\[
\begin{bmatrix}
1 & 0 & 0 & -2 & 3 \\
0 & 1 & 3 & 2 & 0 \\
0 & 0 & 1 & 1 & 0 \\
0 & 0 & 0 & 0 & 0
\end{bmatrix}
\]
→
\( R_2 \to R_2 - 3R_3 \)
giving the reduced row echelon form
\[ A \approx \begin{bmatrix} 1 & 0 & 0 & -2 & 3 \\ 0 & 1 & 0 & -1 & 0 \\ 0 & 0 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 \end{bmatrix} \]Step 2 — Identify pivot columns
Pivot columns: \( x_1, x_2, x_3 \) | Free variables: \( x_4, x_5 \) ⇒ let \( x_4 = t,\; x_5 = s \)
Step 3 — Write equations from the row‑echelon form
\[ \begin{aligned} x_1 - 2x_4 + 3x_5 &= 0 \quad (1) \\ x_2 - x_4 &= 0 \quad (2) \\ x_3 + x_4 &= 0 \quad (3) \end{aligned} \]From (1): \( x_1 - 2t + 3s = 0 \;\Rightarrow\; x_1 = 2t - 3s \)
From (2): \( x_2 - t = 0 \;\Rightarrow\; x_2 = t \)
From (3): \( x_3 + t = 0 \;\Rightarrow\; x_3 = -t \)
Step 4 — General solution
\[ \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \\ x_5 \end{bmatrix} = \begin{bmatrix} 2t - 3s \\ t \\ -t \\ t \\ s \end{bmatrix} = t\begin{bmatrix} 2 \\ 1 \\ -1 \\ 1 \\ 0 \end{bmatrix} + s\begin{bmatrix} -3 \\ 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} \]Step 5 — Basis and dimension
Basis of the solution space:
\[ \mathcal{B} = \left\{ \begin{bmatrix} 2 \\ 1 \\ -1 \\ 1 \\ 0 \end{bmatrix},\; \begin{bmatrix} -3 \\ 0 \\ 0 \\ 0 \\ 1 \end{bmatrix} \right\} \]Dimension of the solution space: \( \dim = 2 \)
Note: this worked solution reduces \(A\) to solve \(Ax = 0\), producing a basis for the
null space of the coefficient matrix built from \(v_1, v_2, v_3, v_4\). If your goal
is instead a basis for the space actually spanned by \(v_1, v_2, v_3, v_4\) (i.e. the
row space), take the nonzero rows of the row‑echelon form of \(A\) as the basis vectors.
Tutorial Video: Basis & Dimension of Homogeneous System
No comments:
Post a Comment